Assume the snow is uniform, and horizontal.
Given:
coefficient of kinetic friction = 0.10 = muK
weight of sled = 48 N
weight of rider = 660 N
normal force on of sled with rider = 48+660 N = 708 N = N
Force required to maintain a uniform speed
= coefficient of kinetic friction * normal force
= muK * N
= 0.10 * 708 N
=70.8 N
Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.
Answer:
<em>20cm</em>
Explanation:
<em>A coil wire comprises of 5 turns, it has a current of 2mA </em>
<em>A net magnetic field at the center of B is equal to Mu over 40 end fractions times T</em>
<em>The radius of a coil of wire with N turns is r = 0.28 m. A clockwise current of Icoil = 1.0 A flows in the coil,</em>
<em>Therefore 40/2mA = 20</em>
Gravitational force between 2 objects . . .
F = G · m₁ · m₂ / D²
-- You said that F = 3.5 x 10²² Newtons.
-- G = the gravitational constant = 6.67 x 10⁻¹¹ N m² / kg²
-- You want to find D .
F = G · m₁ · m₂ / D²
Multiply each side by D² . . . D² · F = G · m₁ · m₂
Divide each side by F . . . D² = G · m₁ · m₂ / F
So finally . . . D = √(G · m₁ · m₂ / F )
D = √(6.67 x 10⁻¹¹ N·m²/kg² · Earth mass · Sun mass / 3.5 x 10²² N)
<em>D = 4.37 x 10⁻¹⁷ · √(Earth mass · Sun mass) </em> meters